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New answer posted
2 months agoContributor-Level 10
We know
Energy on any orbit is given by,
[n1 = 1] (1)
For 3rd orbit
[n2 = 3]
E3 = 13.6ev
= 13.6 – (13.6 * 9)
= 114.15 * 1010m
New answer posted
2 months agoContributor-Level 10
According to Rutherford, e- revolves around in nucleus in circular orbit. Thus e- is always accelerating (centripetal acceleration). An accelerating change emits EM radiation and thus e- should loose energy and finally should collapse in the nucleus.
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